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This one follows on from the question I just asked about logarithms.. Turns out 1/x questions confuse me (sorry for bombarding your exchange with questions, this isn't homework or anything I am just trying to understand stuff)

Okay so I am trying to work out the phase of the following complex number 1/jw

My usual method would be to plot this on an Argand diagram and then take the inverse tangent to get the angle

So as an example if the number was 5+2jw this would be tan^-1(imag/real) = tan^-1(2/5)

However with 1/jw there is no real and the imaginary is 1/jw so I would attempt the same thing imag/real which is (1/jw)/0 =infinity.. Tan^-1(infinity) =90.. So why does Matlab tell me the answer is -90??

Ps I am very sorry for not using formatting I am on my ipad and also am not sure how to do it..

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Take into acount that:

$$z = 1/j = j/j^2 = - j = 1|_{-\pi/2}.$$

If you actually want to obtain the phase of the complex number through the "tangent method", then:

$$\alpha = \arctan(\text{Im}(z)/\text{Re}(z)) = \arctan(-\infty) = - \pi/2.$$

Cheers!

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  • $\begingroup$ Hi thank you for your response.. Please could you explain the pi/2 bit.. I understand that you have equated 1/j to j/j^2 and j^2 =-1 but then I get a bit lost.. $\endgroup$
    – Shasam
    Commented Jan 16, 2014 at 11:47
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    $\begingroup$ Of course. When considering a complex number in cartesian form, say $z = a + b i$, it can be represented in polar form as follows: $z = M|_{\theta}$ where $M = \sqrt{a^2+b^2}$ is its modulus and $\theta = \arctan{b/a}$ is its argument. For example $1+i = \sqrt{2}|_{\pi/4}$. Cheers! $\endgroup$
    – Dmoreno
    Commented Jan 16, 2014 at 11:52
  • $\begingroup$ Ahh right it was just a different kind of notation. I get it now.. There are too many ways to represent a complex number in my opinion! Thanks a Lot $\endgroup$
    – Shasam
    Commented Jan 16, 2014 at 12:28

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